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Chemistry

Chemistry. Session. COORDINATION COMPOUNDS AND ORGANOMETALLICS - 1. Session Objectives. Werner’s coordination theory IUPAC nomenclature Isomerism in Coordination Compounds. Coordination compounds. Coordination compound, central metal ion, ligand, coordination sphere, coordination number.

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Chemistry

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  1. Chemistry

  2. Session COORDINATION COMPOUNDS AND ORGANOMETALLICS - 1

  3. Session Objectives • Werner’s coordination theory • IUPAC nomenclature • Isomerism in Coordination Compounds

  4. Coordination compounds Coordination compound, central metal ion, ligand, coordination sphere, coordination number

  5. Werner’s coordination theory Metal has two types of valencies. • 1o valency - ionisable, non-directional, equal to the valency of the metal. • 2° valency - non-ionisable, directional, equal to the coordination number- decides the hybridization and geometry of molecule.

  6. Illustrative Example An isomer of CrCl3 6H2O reacts with one mole AgNO3 to give one mole of precipitate of AgCl. What is the compound? Solution Only those ions which are present outside the coordination sphere can be ionised. [CrCl2(H2O)4]Cl . H2O + AgNO3  AgCl Therefore the compound is [CrCl2(H2O)4]Cl.H2O.

  7. Basic terms used in coordination chemistry Coordination entity: it constitutes a central atom/ion, usually of a metal, to which are attached a fixed number of other atoms or groups each of which is called a ligand. For example [PtCl4]2- and [Fe(CN)6]3-. Central atom/ion: in a coordination entity the atom/ion to which are bound a fixed number of ligands in a definite geometrical arrangement around it. For example: Ni2+ in [NiCl2(OH2)4], Fe3+ in [Fe(CN)6]3-. Coordination number: it is determined by the sigma bonds between the ligands and the central atom/ion. For example coordination number in [Fe(CN)6]3- is six.

  8. Oxidation number of the centralmetal ion. [Co(NH3)4 Cl (NO2)]K3[Fe(CN)6] x + 4(o) + 1 (–1) + 1 (–1) = 0 3(1) + x + 6(–1) = 0 x = + 2 x = + 3.

  9. Ligands The ligands are the ions or molecules bound to the central atom/ion in the coordination entity. Ligands are Lewis base because they works as electron donor. Ambidentate ligands: Ligands which can ligate through two different atoms present in it. For example SCN- ion can coordinate through the sulphur or nitrogen atom. Denticity and chelation: when coordination of more than one sigma electron pair donor group from the ligand to the same central atom/ion takes place, it is called chelation. The number of such ligating groups indicate the denticity of the ligand, e.g., ethylenediamine, EDTA For example unidentate, didentate, terdentate tetradentate etc.

  10. Some common ligands

  11. IUPAC nomenclature • Cation is named first. • Name the coordination sphere. • Ligands are named in the alphabetical orderirrespective of their being neutral, negativelyor positively charged.[Pt Cl (NO2) (NH3)4] amine, chloro, nitrito – N • Numerical prefixes to indicate the number of ligands. –di, tri, etc.

  12. IUPAC nomenclature • Ending form: For anionic complex,metal ion ends with ‘ate’. • K[Pt Cl5 (NH3)] potassium ammine pentachloro-platinate (IV) • Latin name of the metals are commonly used.cuperate for Cu, ferrate for Fe, argentate for Ag. • Oxidation state of metal is designated by a Romannumerical. (Such as II, III, IV) • Bridging of ligand is indicated with a prefix-.

  13. IUPAC nomenclature [Co(NH3)5(NCS)]Cl2 Pentaammineisothiocyanatocobalt (III) chloride [Cr(H2O)4Cl2]+ Tetraaquadichlorochromium (III) ion K2HgI4 Potassium tetraiodomercurate (II) Sodium tetrafluorooxochromate (IV) Na2[CrOF4] Bis(ethylenediamine)cobalt (III)- m - imido - m – hydroxo bis(ethylenediammine)cobalt (III) ion.

  14. Rules for writing formula of complexes • complexes are enclosed in square brackets. • first the name of the central atom is given. • followed by first the anionic ligand and then the neutral ligands; within each group they are alphabetically ordered according to the first character of their formula. • Examples: [PtCl2(C2H4)(NH3)] K2[PdCl4] [Co(en)3]Cl3

  15. Illustrative example • Write the oxidation state of central metal atom in the following compounds. • Ag in Tollen’s reagent is • [Mo2O4(H2O)2]2– Solution: (i) Ag in Tollen’s reagent exists as Ag2O 2x + 1(–2) = 0 x = +1 (ii) 2x + 4(–2) + 2 × 0 = –2 x = +3

  16. Illustrative Example • Write IUPAC name of the following compounds: • K2[Zn(OH)4] • [CoCl.CN.NO2.(NH3)3] • [Pt(py)4][PtCl4] • K2[Cr(CN)2O2(O2)NH3] Solution : Follow IUPAC rules. • Potassium tetrahydroxozincate(II) • Triamminechlorocyanonitrocobalt(III) • Tetrapyridineplatinum(II) tetrachloroplatinate(II) • Potassium amminedicyanodioxoperoxochromate(VI)

  17. Illustrative Example Write the formula of the following coordination compounds. i. Bis(acetylacetonato) oxovanadium (IV) ii. Pentaammine carbonatocobalt (III) chloride iii. Sodium tetratacyanonitrosylsulphidoferrate (III) iv. Dichlorobis (urea) copper (II) Solution: i. [VO(acac)2] ii. [Co(NH3)5CO3]Cl iii. Na3[Fe(CN)4NOS] iv. [Cu{(NH2)2CO}2Cl2]

  18. Effective atomic number(EAN) Atomic number of cobalt is 27.As such Co3+ has 24 electrons. Each of the six NH3 will donate one pair of electron to cobalt. So EAN of Co3+ is 24 + 12 = 36.

  19. Illustrative Example • What is the EAN of the central atom in following compounds? • K3[Fe(CN)6] • K4[Fe(CN)6] • [Cu(NH3)4]SO4 Solution : • EAN of [Fe(CN)6]3- • 26 – 3 + 2 × 6 = 35 (2) EAN of [Fe(CN)6]4– 26 – 2 + 2 × 6 = 36 (3) EAN of [Cu(NH3)4]2– 29 – 2 + 2 × 4 = 35

  20. [Co(NH3)5SO4] Br Isomerism in Coordination Compounds (i) Structural isomerism (ii) Stereo-isomerism Structural isomerism Ionization isomerism

  21. Isomerism Hydrate isomerism [Cr(H2O)6]Cl3 (Violet) [Cr(H2O)5Cl]Cl2.H2O (Green) [Cr(H2O)4Cl2]Cl.2H2O (Green) Coordination isomerism [Co(NH3)6] [Cr (CN)6] and [Cr(NH3)6] [Co (CN)6]

  22. Isomerism Linkage isomerism This type of isomerism occurs in ambidentate ligands like and CO. Example [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2 Coordination position isomerism

  23. Illustrative Example Write the type of isomerism in the compounds [Co(NH3)4Cl2]NO2 and [Co(NH3)4ClNO2]Cl. Solution: As the two complexes give different ions in solution, they show ionization isomerism.

  24. Geometrical isomerism CN=4 Complexes with general formula, Ma2b2

  25. Geometrical isomerism Complexes with general formula Ma2bc can have cis- and trans-isomers.

  26. Geometrical isomerism

  27. Geometrical isomerism Complexes with general formula, Mabcd, can have three isomers.

  28. Geometrical isomerism CN=6 Octahedral complexes of the type Ma4b2 Trans-isomer Cis-isomer

  29. Geometrical isomerism

  30. Geometrical isomerism Octahedral complexes of the type Ma3b3

  31. Geometrical isomerism three NH3 molecules can be situated (1) on the same face of the octahedron (fac isomer) or (2) around a perimeter of the octahedron (mer isomer

  32. Optical isomerism

  33. Optical isomerism (Where AA = Symmetrical bidentate ligand) (Where AB = Unsymmetrical ligands)

  34. Illustrative Example A metal complex having composition Cr(NH3)4Cl2Br has been isolated in two forms (A) and (B). The form (A) reacts with AgNO3 to give a white precipitate readily soluble in dilute aqueous ammonia whereas (B) gives a pale yellow precipitate soluble in concentrated ammonia. Write the formula of (A) and (B) and state the hybridisation of chromium in each. Calculate the magnetic moment (spin only value). Solution : Since A is readily soluble in dilute aqueous ammonia. It must have Cl-outside the coordination sphere I.e. its formula is [Cr(NH3)4ClBr] Cl Similarly, form B gives pale yellow precipitate of AgBr which are sparingly soluble in NH4OH.

  35. Solution Cont. The state of hybridisation of chromium in both the complexes is d2sp3. Both the forms have chromium in +3 oxidation state. As there are three unpaired electrons so magnetic moment

  36. Platinum II forms square planar complexes and platinum IV gives octahedral complexes. How many geometrical isomers are possible for each of the following complexes. Describe their structures. Illustrative Example Solution (a) No isomers are possible for a square planar complex of the type MA3B

  37. Solution contd. (b) No isomers are possible for an octahedral complex of the type of MAB5. (c). Cis and trans isomers are possible for a square planar complex of the type MA2BC.

  38. Solution Contd. (d). Cis and trans isomers are possible for an octahedral complex of the type MA4BC.

  39. Thank you

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